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                [Ncpc2010]Around the track                </h1>
                <p>时间限制：1s&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  空间限制：128MB</p>			</div>
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                                <h3>题目描述</h3><p><p>给你一个图。你找一条欧拉回路出来，使得沿这条路转过的角度最小。<br />
图很特殊，一个结点出来，要么恰好有2条边，或者恰好有4边<br />
一个欧拉回路，即它必须遍历每个图形的边恰好一次，并回到起 点。 （输入的跑道保证其欧拉回路的存在）.总的转弯弧度就是在这个回路中， 在每个结点处需要的转弯弧度的总和，在一条直线上继续行进不用转弯。每一段 跑道都是可以双向行驶的。</p>
<p></p></p><hr/><h3>输入格式</h3><p><p>Oneline with 3 &lt; =&nbsp; N &lt; =&nbsp; 10000{the number of nodes{and N &lt; =&nbsp; M &lt; =&nbsp; 2N {the number<br />
of edges.<br />
N lines with the x and y coordinates o feachnode,inorder.0 &lt; =&nbsp; x;y &lt; =&nbsp; 10000.The nodes<br />
have unique coordinate pairs.<br />
M lines with two space separated numbers i and j,denoting an edge between nodes i and<br />
j.The nodes are 0-indexed.</p>
<p>第一行 两个数,N , M . (3&lt;=N &lt;=10000, 代表结点的个数。N&lt;=M&lt;=2N 边的个数)<br />
接下来N行，依次组出每个结点的坐标&nbsp; 0&lt;=x,y&lt;=10000 。每个结点的坐标是唯一 的<br />
接下来M行,每两个数，表示i, j之间存在一条边.(结点从0开始编号)</p>
<p></p></p><hr/><h3>输出格式</h3><p><p>欧拉回路上最小的所需转弯的弧度和 <br />
</p></p><hr/><h3>样例输入</h3><pre>12 19
1077 2677
7473 4262
1095 8844
84 7875
7241 7320
9143 4888
4524 1947
4652 1260
3503 7882
4692 223
9745 5245
2037 2387
0 1
0 2
1 4
1 2
1 3
3 4
4 6
4 5
5 8
5 6
5 7
6 8
6 7
7 9
7 8
8 9
9 11
9 10
10 11
</pre><hr/><h3>样例输出</h3><pre>37.699111843077446
</pre><hr/><h3>提示</h3><p>没有写明提示</p><hr/><h3>题目来源</h3><p>没有写明来源</p>
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